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Stels [109]
3 years ago
12

I need the answer to this asap!! please explain

Chemistry
1 answer:
kumpel [21]3 years ago
6 0

Answer:

See explanation

Explanation:

Calcium carbide reacts with water to yield acetylene gas and calcium hydroxide as follows;

CaC2(s)+2H2O(g)⇋Ca(OH)2(s)+C2H2(g)

This now shows us that the equation as written in the question is wrong. Since the equation for the reaction of calcium carbide and water as shown in the question is wrong, the equation can not be balanced.

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The______ the forces among the particles in a sample of matter, the more rigid the matter will be.
Alexus [3.1K]

Answer:

Stronger!

Explanation:

The <u>stronger</u> the forces among the particles in a sample of matter, the more rigid the matter will be.

3 0
3 years ago
How many grams of a stock solution that is 87.5 percent H2SO4 by mass would be needed to make 275 grams of a 55.0 percent by mas
dalvyx [7]
87+55=142
5+0=5
The answer is 142.5
8 0
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HELP QUICK!! Match the term with the definition. (4 points)
Rufina [12.5K]

Answer:

1. c.

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3 years ago
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The effectiveness of nitrogen fertilizers depends on both their ability to deliver nitrogen to plants and the amount of nitrogen
densk [106]

Answer:

Ammonia > Urea > Ammonium nitrate > Ammonium sulphate

Explanation:

Percentage by mass of nitrogen in NH3:

Molar mass of NH3= 17 g/mol

Hence % by mass = 14/17 × 100 = 82.35%

% by mass of NH4NO3

Molar mass of NH4NO3 = 80.043 g/mol

Hence; 28/80.043 × 100 = 34.98%

% by mass of (NH4)2SO4;

Molar mass of (NH4)2SO4= 132.14 g/mol

Hence; 28/132.14 × 100 = 21.19%

% by mass of CH4N2O

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Hence 28/60.0553 × 100 = 46.62%

8 0
3 years ago
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
vodomira [7]

Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
  • pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

3 0
3 years ago
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